In the following nuclear reaction $x$ stands for $n \rightarrow p+e^{-}+x$

In the following nuclear reaction $x$ stands for $n \rightarrow p+e^{-}+x$
  1. $\alpha$-particle
  2. positron
  3. nutrino
  4. antinutrino

Solution

Here $\frac{n}{p} \downarrow \downarrow$. So it is $\beta^{-}$decay. So, $x$ stands for antinutrino.

Asked in: AP EAMCET 2015

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